Graph y=x^2+4x+7

Problem

y=x2+4*x+7

Solution

1. Identify the type of function and its orientation. This is a quadratic function in the form y=a*x2+b*x+c Since a=1 which is positive, the parabola opens upward.

2. Find the x-coordinate of the vertex using the formula x=(−b)/(2*a)

x=(−4)/(2*(1))

x=−2

3. Find the y-coordinate of the vertex by substituting x=−2 into the original equation.

y=(−2)2+4*(−2)+7

y=4−8+7

y=3

The vertex is at (−2,3)

4. Determine the y-intercept by setting x=0

y=(0)2+4*(0)+7

y=7

The y-intercept is (0,7)

5. Find additional points to define the shape. Since the axis of symmetry is x=−2 the point (0,7) can be reflected across the axis to find the point (−4,7)

6. Check for x-intercepts by setting y=0

0=x2+4*x+7

Using the discriminant D=b2−4*a*c

D=4−4*(1)*(7)

D=16−28

D=−12

Since the discriminant is negative, there are no real x-intercepts.

Final Answer

To graph y=x2+4*x+7 plot the vertex at (−2,3) the y-intercept at (0,7) and its symmetric point at (−4,7) then draw a smooth upward-opening parabola through these points.

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